I'm working on the following exercise:
"Let $U=\{p\in P_4(\mathbb{F}):p(6)=0\}$. Find a basis of $U$."
What I've done:
$p\in U\iff p=c_0+c_1 x+c_2 x^2+c_3x^3+c_4 x^4$ and $p(6)=0$ $\iff c_0+6c_1+36c_2+216c_3+1296c_4=0 \iff (c_0,c_1,c_2,c_3,c_4)=a(6,-1,0,0,0)+b(0,6,-1,0,0)+c(0,0,6,-1,0)+d(0,0,0,6,-1); a,b,c,d\in\mathbb{F}$
so a basis of $U$ is $\{6-x,6x-x^2,6x^2-x^3,6x^3-x^4\}$.
Is this right? Is there a better way to write up/justify this result? Is there another (perhaps simpler) way to arrive at this result? I'd like to know. Thanks.
Yes it is correct, as a quick alternative we can easily find it by choosing: $$(x-6),(x-6)^2,(x-6)^3,(x-6)^4$$
Indeed we have 4 vectors and
$$p=c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4=0$$
is true if and only if $p$ is the zero polynomial with $c_i=0$ $\forall i$ then they are linearly independent.